Python Dictionaries: Medical Insurance Project

[suzanne792]

Hi everyone:
I’ve seen this question asked before, but the answer was, "actually, that’s not the result you’re getting. But… it is! When I run the code shown below, print(names_to_ages) returns “{}” instead of the populated dictionary.

Can anyone please tell me what I’m doing wrong? I got frustrated with it and copied the code directly from the “hint” text, but i still get the same answer. Tried copying from the solutions too, but had the same problem.

names = [“Marina”, “Vinay”, “Connie”, “Isaac”, “Valentina”]

ages = [27, 24, 43, 35, 52]

zipped_ages = zip(names, ages)

print (“Zipped:”, list(zipped_ages))

print(“”)

names_to_ages = {key: value for key, value in zipped_ages}

print(names_to_ages)

My results are as follows:
Zipped: [(‘Marina’, 27), (‘Vinay’, 24), (‘Connie’, 43), (‘Isaac’, 35), (‘Valentina’, 52)]

{}

Please, please, can anyone tell me what I’ve done wrong?

[suzanne792]

And, I figured out how to fix it, but I don’t know why it was a problem.

If I remove the line: print(“Zipped:”, list(zipped_ages)) it all works. Why?

[lisalisaj]

@suzanne792

zipped_ages = zip(names, ages)

zip results in an iterable (tuples to be exact). and, to quote the documentation,
(Built-in Functions — Python 3.12.0 documentation), ‘zip is lazy: The elements won’t be processed until the iterable is iterated on, e.g. by a for loop or by wrapping in a [list] (bold by me) (Built-in Types — Python 3.12.0 documentation).’

I think another way to put it is that zip consumes itself when it combines the two lists (or iterables) and returns a zip object, which is why when you go to print it, you get something like:

zipped_ages = zip(names, ages)

print(zipped_ages)
<zip object at 0x7ff208c2d080>

So, when you go to use it a 2nd time, here" names_to_ages = {key: value for key, value in zipped_ages}

it’s already been “consumed” with the original zip() and there’s nothing to iterate over. So, you get nothing, or an empty dictionary. You have to cast it as a list().

We could tweak your code:

names = ["Marina", "Vinay", "Connie", "Isaac", " Valentina"]

ages = [27, 24, 43, 35, 52]

zipped_ages = zip(names, ages)

print(list(zipped_ages))

>>>[('Marina', 27), ('Vinay', 24), ('Connie', 43), ('Isaac', 35), (' Valentina', 52)]


#and, make it:
names = ["Marina", "Vinay", "Connie", "Isaac", " Valentina"]

ages = [27, 24, 43, 35, 52]

zipped_ages = list(zip(names, ages))

print(zipped_ages)

>>>[('Marina', 27), ('Vinay', 24), ('Connie', 43), ('Isaac', 35), (' Valentina', 52)]

#and then, use your code:
names_to_ages = {key: value for key, value in zipped_ages}

print(names_to_ages)

>>>{'Marina': 27, 'Vinay': 24, 'Connie': 43, 'Isaac': 35, ' Valentina': 52}

You could also do something like:

zipped_ages = dict(zip(names, ages))
print(zipped_ages)

>>>{'Marina': 27, 'Vinay': 24, 'Connie': 43, 'Isaac': 35, ' Valentina': 52}

But, I don’t think that’s what the lesson wants you to do. (I cannot recall).

Did you have a look at this topic? it might be helpful.

Zip creates an iterator, how can we turn that into a dictionary?

why the dictionary is empty when i execute this lines of code? drinks = ["espresso", "chai", "chai", "drip"] caffeine = [64, 40, 0, 120] zipped_drinks = zip(drinks, caffeine) print("Zipped drinks ",list(zipped_drinks)) drinks_to_caffeine = {key: value for key, value in zipped_drinks} print("Drinks to caffeine dictionary ",drinks_to_caffeine) output: Zipped drinks [(‘espresso’, 64), (‘chai’, 40), (‘chai’, 0), (‘drip’, 120)] Drinks to caffeine dictionary {}

[lisalisaj]

because you’re iterating over the zipped object with a for loop.

print(zipped_ages) #prints nothing

names_to_ages = {key: value for key, value in zipped_ages} #this works b/c you're iterating over the zipped object

>>><zip object at 0x7ff208c2dfc0>
{'Marina': 27, 'Vinay': 24, 'Connie': 43, 'Isaac': 35, ' Valentina': 52}

[suzanne792]

Thank you! The problem came when I tried to print list(zipped_ages), but it worked as soon as I removed that line. I like the idea of setting it equal to a list of the zipped_ages when zipping them. I tried it in my code and it worked perfectly.

You rock!!

[lisalisaj]

Sure thing.

This: print(list(zipped_ages) doesn’t change the original zipped_ages. It just changes it in that piece of code (for lack of a better phrase. Sorry, I’m tired. lol).

So, when you comment out that piece, you end up iterating over that zipped object with your for loop and create the dictionary.